Prove $h^{-1}fh in K_4$ for $fin K_4$, $hin S_4$.
$begingroup$
I want to show if $f in K_4={id,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}$, then for all permutations $h in S_4$ also $h^{-1}fh in K_4$ (not using group theory).
The elements in $K_4$ except for the identity are the elements in $S_4$ which are a product of two disjoint 2-cycles.
Therefore, I am trying to prove that $h^{-1}fh$ in disjoint cycle notation are two 2-cycles.
Then I consider different cycle structures of $h$, $4$ 1-cycles, one 3-cycle and one 1-cycle etc.
However, I am not sure why it is true for instance if $h$ is $(1)(2 3 4)$, taking the $f$ to be the identity I get $(1)(2)(3 4)$ which is not two 2-cycles.
group-theory permutations
edited Dec 22 '18 at 2:36
Shaun
9,804113684
asked Nov 23 '16 at 15:28
user30523user30523
564510
$endgroup$
add a comment |
$begingroup$
I want to show if $f in K_4={id,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}$, then for all permutations $h in S_4$ also $h^{-1}fh in K_4$ (not using group theory).
The elements in $K_4$ except for the identity are the elements in $S_4$ which are a product of two disjoint 2-cycles.
Therefore, I am trying to prove that $h^{-1}fh$ in disjoint cycle notation are two 2-cycles.
Then I consider different cycle structures of $h$, $4$ 1-cycles, one 3-cycle and one 1-cycle etc.
However, I am not sure why it is true for instance if $h$ is $(1)(2 3 4)$, taking the $f$ to be the identity I get $(1)(2)(3 4)$ which is not two 2-cycles.
group-theory permutations
edited Dec 22 '18 at 2:36
Shaun
9,804113684
asked Nov 23 '16 at 15:28
user30523user30523
564510
$endgroup$
$begingroup$
You made a mistake in the example you mention. The result is the identity.
$endgroup$
– Tobias Kildetoft
Nov 23 '16 at 15:35
add a comment |
$begingroup$
I want to show if $f in K_4={id,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}$, then for all permutations $h in S_4$ also $h^{-1}fh in K_4$ (not using group theory).
The elements in $K_4$ except for the identity are the elements in $S_4$ which are a product of two disjoint 2-cycles.
Therefore, I am trying to prove that $h^{-1}fh$ in disjoint cycle notation are two 2-cycles.
Then I consider different cycle structures of $h$, $4$ 1-cycles, one 3-cycle and one 1-cycle etc.
However, I am not sure why it is true for instance if $h$ is $(1)(2 3 4)$, taking the $f$ to be the identity I get $(1)(2)(3 4)$ which is not two 2-cycles.
group-theory permutations
edited Dec 22 '18 at 2:36
Shaun
9,804113684
asked Nov 23 '16 at 15:28
user30523user30523
564510
$endgroup$
I want to show if $f in K_4={id,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}$, then for all permutations $h in S_4$ also $h^{-1}fh in K_4$ (not using group theory).
The elements in $K_4$ except for the identity are the elements in $S_4$ which are a product of two disjoint 2-cycles.
Therefore, I am trying to prove that $h^{-1}fh$ in disjoint cycle notation are two 2-cycles.
Then I consider different cycle structures of $h$, $4$ 1-cycles, one 3-cycle and one 1-cycle etc.
However, I am not sure why it is true for instance if $h$ is $(1)(2 3 4)$, taking the $f$ to be the identity I get $(1)(2)(3 4)$ which is not two 2-cycles.
group-theory permutations
group-theory permutations
edited Dec 22 '18 at 2:36
Shaun
9,804113684
asked Nov 23 '16 at 15:28
user30523user30523
564510
edited Dec 22 '18 at 2:36
Shaun
9,804113684
asked Nov 23 '16 at 15:28
user30523user30523
564510
edited Dec 22 '18 at 2:36
Shaun
9,804113684
edited Dec 22 '18 at 2:36
Shaun
9,804113684
edited Dec 22 '18 at 2:36
Shaun
9,804113684
9,804113684
asked Nov 23 '16 at 15:28
user30523user30523
564510
asked Nov 23 '16 at 15:28
user30523user30523
564510
asked Nov 23 '16 at 15:28
user30523user30523
564510
564510
$begingroup$
You made a mistake in the example you mention. The result is the identity.
$endgroup$
– Tobias Kildetoft
Nov 23 '16 at 15:35
add a comment |
$begingroup$
You made a mistake in the example you mention. The result is the identity.
$endgroup$
– Tobias Kildetoft
Nov 23 '16 at 15:35
$begingroup$
You made a mistake in the example you mention. The result is the identity.
$endgroup$
– Tobias Kildetoft
Nov 23 '16 at 15:35
$begingroup$
You made a mistake in the example you mention. The result is the identity.
$endgroup$
– Tobias Kildetoft
Nov 23 '16 at 15:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What you're trying to show is that $K_4$ is a normal subgroup of $S_4$.
However, you said you wish not to use group theory. No problem.
Theorem: For any $ninBbb N$ and permutations $alpha,betain S_n$ with
$$beta=prod_{i}(b_{i,1}cdots b_{i,j_i}),$$
we have that
$$alpha^{-1}betaalpha=prod_{i}(alpha(b_{i,1})cdots alpha(b_{i,j_i})),tag{1}$$
where $alpha(x)$ is $alpha$ applied to $xin overline{1,n}:={1,dots, n}$, and the cycle structure of $beta$ is preserved in the RHS of $(1)$.
Then one just needs to observe that each $hin S_4$ is a bijection on $overline{1,4} $, so that the disjoint cycle nature of each $f$ in $K_4$ is preserved.
And as Tobias pointed out, you made an error in your calculation.
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you're trying to show is that $K_4$ is a normal subgroup of $S_4$.
However, you said you wish not to use group theory. No problem.
Theorem: For any $ninBbb N$ and permutations $alpha,betain S_n$ with
$$beta=prod_{i}(b_{i,1}cdots b_{i,j_i}),$$
we have that
$$alpha^{-1}betaalpha=prod_{i}(alpha(b_{i,1})cdots alpha(b_{i,j_i})),tag{1}$$
where $alpha(x)$ is $alpha$ applied to $xin overline{1,n}:={1,dots, n}$, and the cycle structure of $beta$ is preserved in the RHS of $(1)$.
Then one just needs to observe that each $hin S_4$ is a bijection on $overline{1,4} $, so that the disjoint cycle nature of each $f$ in $K_4$ is preserved.
And as Tobias pointed out, you made an error in your calculation.
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
$endgroup$
add a comment |
$begingroup$
What you're trying to show is that $K_4$ is a normal subgroup of $S_4$.
However, you said you wish not to use group theory. No problem.
Theorem: For any $ninBbb N$ and permutations $alpha,betain S_n$ with
$$beta=prod_{i}(b_{i,1}cdots b_{i,j_i}),$$
we have that
$$alpha^{-1}betaalpha=prod_{i}(alpha(b_{i,1})cdots alpha(b_{i,j_i})),tag{1}$$
where $alpha(x)$ is $alpha$ applied to $xin overline{1,n}:={1,dots, n}$, and the cycle structure of $beta$ is preserved in the RHS of $(1)$.
Then one just needs to observe that each $hin S_4$ is a bijection on $overline{1,4} $, so that the disjoint cycle nature of each $f$ in $K_4$ is preserved.
And as Tobias pointed out, you made an error in your calculation.
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
$endgroup$
add a comment |
$begingroup$
What you're trying to show is that $K_4$ is a normal subgroup of $S_4$.
However, you said you wish not to use group theory. No problem.
Theorem: For any $ninBbb N$ and permutations $alpha,betain S_n$ with
$$beta=prod_{i}(b_{i,1}cdots b_{i,j_i}),$$
we have that
$$alpha^{-1}betaalpha=prod_{i}(alpha(b_{i,1})cdots alpha(b_{i,j_i})),tag{1}$$
where $alpha(x)$ is $alpha$ applied to $xin overline{1,n}:={1,dots, n}$, and the cycle structure of $beta$ is preserved in the RHS of $(1)$.
Then one just needs to observe that each $hin S_4$ is a bijection on $overline{1,4} $, so that the disjoint cycle nature of each $f$ in $K_4$ is preserved.
And as Tobias pointed out, you made an error in your calculation.
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
$endgroup$
What you're trying to show is that $K_4$ is a normal subgroup of $S_4$.
However, you said you wish not to use group theory. No problem.
Theorem: For any $ninBbb N$ and permutations $alpha,betain S_n$ with
$$beta=prod_{i}(b_{i,1}cdots b_{i,j_i}),$$
we have that
$$alpha^{-1}betaalpha=prod_{i}(alpha(b_{i,1})cdots alpha(b_{i,j_i})),tag{1}$$
where $alpha(x)$ is $alpha$ applied to $xin overline{1,n}:={1,dots, n}$, and the cycle structure of $beta$ is preserved in the RHS of $(1)$.
Then one just needs to observe that each $hin S_4$ is a bijection on $overline{1,4} $, so that the disjoint cycle nature of each $f$ in $K_4$ is preserved.
And as Tobias pointed out, you made an error in your calculation.
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
edited Dec 22 '18 at 8:05
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
answered Dec 22 '18 at 2:33
ShaunShaun
9,804113684
9,804113684
add a comment |
add a comment |
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$begingroup$
You made a mistake in the example you mention. The result is the identity.
$endgroup$
– Tobias Kildetoft
Nov 23 '16 at 15:35